Do analytic continuations exist over $\mathbb{R}$?

Question

I have a feeling this is quite obvious since $\mathbb{R}\subset\mathbb{C}$, but suppose we have the function $f:\left[-1,1\right]\times\mathbb{R}\to\mathbb{R}$ defined by

$$f\left(x,y\right) = \frac{y\left(x-\mathrm{csch}\,y\,e^{xy}+\coth y\right)}{y\coth y - 1 - \log\left(y\,\mathrm{csch}\, y\right)}$$

It is easy to see that $f$ is undefined for $y=0$. But, from my understanding, $y=0$ is, in fact, a removable singular point for $f$ since

$$\lim_{y\to 0} f\left(x,y\right) = 1-x^2$$

which one can determine by considering the appropriate Laurent expansion. I am wondering if it is accurate to claim that if I assign $f\left(x,y=0\right) = 1-x^2$, then I have constructed an analytic continuation of $f$ that is valid over all $y\in\mathbb{R}$ (and, of course, $x\in\left[-1,1\right]$). Any help is appreciated!

Answer 1

No—rather, such a smooth continuation is not unique and therefore not worth calling an "analytic continuation." The values of a smooth function on an interval do not determine the values of the function elsewhere. For example, for any smooth function $f\colon[a,b]\to\Bbb R$, there is a "real-analytic continuation" $g\colon\Bbb R\to\Bbb R$ which

• is smooth,
• agrees with $f$ on $[a-\varepsilon,b+\varepsilon]$ for any given $\varepsilon>0$,
• and for which $g(x)=0$ for $x\in\Bbb R\setminus[a-\varepsilon,b+\varepsilon]$.